What is the solution for y^{''}+y=cos(2x) ?

The solution for y^{''}+y=cos(2x) is y=c_{1}cos(x)+c_{2}sin(x)-1/3 cos(2x)

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y^{''}+y=cos(2x)

y^{^{''}} + y = cos (2 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} cos (2 x)

Solution steps

y^{''} (x) + y = cos (2 x)

Solve linear ODE:

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} cos (2 x)

y = c_{1} cos (x) + c_{2} sin (x) - \frac{1}{3} cos (2 x)

\int (2 - x)^{2} d x

3 y^{^{''}} - 4 y^{^{'}} - 4 y = 0, y (0) = - 7, y^{^{'}} (0) = - 8

\int (e^{x} + x) d x

y = 4 x^{2} + 5 x + 1, (0, 1)

\frac{d}{d x} (lo g_{10} (lo g_{10} (x)))

What is the solution for y^{''}+y=cos(2x) ?
The solution for y^{''}+y=cos(2x) is y=c_{1}cos(x)+c_{2}sin(x)-1/3 cos(2x)